## 数学代写|数学分析代写Mathematical Analysis代考|MATH2050

2023年2月2日

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## 数学代写|数学分析代写Mathematical Analysis代考|Continuous Functions on Compact Sets

A subset $K$ in a metric space $(X, d)$ is compact if every sequence $x_k$ of points of $K$ admits a subsequence that converges to a point $x \in K$.
If $K \subseteq X$ is a compact set according to the above definition, it is usually called sequentially compact in order to distinguish the above from the definition in the Appendix (see Sect. 2.12), which uses open covers. At any rate, in Sect. $2.12$ we shall prove that the metric structure of the space $X$ guarantees that the two notions are equivalent.

Proposition If $(X, d)$ is a metric space and $K \subseteq X$ is compact, then $K$ is closed.
Proof Take a sequence $x_k$ of points in $K$ with limit $x \in X$. We claim $x \in K$. As $K$ is compact, there exists a subsequence $x_{k_n}$ converging to some $x_0 \in K$. Necessarily $x=x_0$, and so $x \in K$.
In case $K=X$ is compact, we call $(X, d)$ a compact metric space.
By relying on the Bolzano-Weierstrass theorem for $\mathbb{R}$, the following result is easy to prove.

Characterisation of Compact Subsets of $\mathbb{R}^n$ (Heine-Borel Theorem) $A$ subset $K \subset \mathbb{R}^n$ is compact if and only if it is closed and bounded.

Proof Suppose $K$ is a closed and bounded subset, and let us take a sequence $x_k=\left(x_{k, 1}, x_{k, 2}, \ldots, x_{k, n}\right)$ in $K$. The sequences $x_{k, i}$, for $i=1,2, \ldots, n$, are bounded sequences of real numbers. The Bolzano-Weierstrass theorem tells that we may extract from $x_k$ a subsequence whose first coordinate converges. From the latter we may extract a subsequence whose second coordinate converges, and so on. This produces a strictly increasing sequence $k_h$ of natural numbers such that $x_{k_h, i} \rightarrow x_i$ for any $i=1,2, \ldots, n$. Therefore $x_{k_{\mathrm{h}}} \rightarrow x=\left(x_1, x_2, \ldots, x_n\right)$. But $K$ is closed, so $x \in K$ and $K$ is compact.

## 数学代写|数学分析代写Mathematical Analysis代考|Connected Open Subsets of Rn

An open subset $A \subseteq \mathbb{R}^n$ is connected when there is no partition $\left{A_1, A_2\right}$ of $A$ into non-empty open sets. Equivalently, an open set $A$ is connected if and only if
$$A_1, A_2 \text { open in } \mathbb{R}^n, \quad A_1 \cap A_2=\emptyset, \quad A_1 \cup A_2=A$$
imply that one of the two sets $A_1$ and $A_2$ is empty.
A segment $\left[x_1, x_2\right]$ in $\mathbb{R}^n$ with endpoints $x_1$ and $x_2$ is the set of points $x(t)$ in $\mathbb{R}^n$ such that
$$x(t)=x_1+t\left(x_2-x_1\right), \quad \forall t \in[0,1] .$$
Let $x_1, x_2, \ldots, x_k$ be $k$ points of $\mathbb{R}^n(k \geq 2)$, with $x_i \neq x_{i+1}$ for any $i=1,2, \ldots, k-1$. A polygonal path in $\mathbb{R}^n$ with vertices $x_1, x_2, \ldots, x_k$ is the union of the line segments $\left[x_i, x_{i+1}\right]$, for $i=1,2, \ldots, k-1$. The points $x_1, x_k$ are the path’s endpoints.
Theorem on Connected Open Subsets of $\mathbb{R}^n$ If $A \subseteq \mathbb{R}^n$ is a connected open set, any pair of points of $A$ are endpoints of a polygonal path entirely contained in $A$.
Proof Pick $x_1 \in A$ and call $A_1$ the set of endpoints $x \in A$ of polygonal paths starting at $x_1$ and all contained in $A$. We claim $A_1=A$.

The set $A_1$ is open, because if $x \in A_1$ then $x$ belongs to the open set $A$ and then there is an open ball $I_\delta(x)$, centred at $x$ with radius $\delta$, contained in $A$. If $y \in I_\delta(x)$, $y \neq x$, the line segment $[x, y]$ is part of a radial segment in the ball $I_\delta(x)$, and is clearly contained in $I_\delta(x)$. Hence joining the previous path between $x_1$ and $x$ with the line segment $[x, y]$ produces a polygonal path from $x_1$ to $y$, still contained in $A$. Therefore $y \in A_1$ by definition, and then $I_\delta(x) \subseteq A_1$, making $A_1$ open.

## 数学代写|数学分析代写Mathematical Analysis代考|Continuous Functions on Compact Sets

Weierstrass 定理告诉我们可以从 $x_k$ 其第一个坐标收敛 的子序列。从后者我们可以提取第二个坐标收敛的子序 列，依此类推。这会产生一个严格递增的序列 $k_h$ 的自然 数使得 $x_{k_h, i} \rightarrow x_i$ 对于任何 $i=1,2, \ldots, n$. 所以 $x_{k_{\mathrm{h}}} \rightarrow x=\left(x_1, x_2, \ldots, x_n\right)$. 但 $K$ 是关闭的，所以 $x \in K$ 和 $K$ 很紧凑。

## 数学代写|数学分析代写Mathematical Analysis代考|Connected Open Subsets of Rn

$A_1, A_2$ open in $\mathbb{R}^n, \quad A_1 \cap A_2=\emptyset, \quad A_1 \cup A_2=$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。